On a smooth inclined plane, a body of mass \(M\) is attached between two springs. The other ends of the springs are fixed to firm supports. If each spring has force constant \(K\), the period of oscillation of the body (assuming the springs as massless) will be:
                
1. \(2\pi \left( \frac{M}{2K}\right)^{\frac{1}{2}}\)
2. \(2\pi \left( \frac{2M}{K}\right)^{\frac{1}{2}}\)
3. \(2\pi \left(\frac{Mgsin\theta}{2K}\right)\)
4. \(2\pi \left( \frac{2Mg}{K}\right)^{\frac{1}{2}}\)

All the surfaces are smooth and the system, given below, is oscillating with an amplitude \({A}.\) What is the extension of spring having spring constant \({k_1},\) when the block is at the extreme position?
              

A spring of force constant \(k\) is cut into lengths of ratio \(1:2:3\). They are connected in series and the new force constant is \(k'\). Then they are connected in parallel and the force constant is \(k''\). Then \(k':k''\) is:
1. \(1:9\)
2. \(1:11\)
3. \(1:14\)
4. \(1:6\)

When a mass is suspended separately by two different springs, in successive order, then the time period of oscillations is \(t _1\) and \(t_2\) respectively. If it is connected by both springs as shown in the figure below, then the time period of oscillation becomes \(t_0.\) The correct relation between \(t_0,\) \(t_1\) & \(t_2\) is:

1. ${{\mathrm{t}}_0}^2={{\mathrm{t}}_1}^2+{{\mathrm{t}}_2}^2$

2. ${{\mathrm{t}}_0}^{-2}={{\mathrm{t}}_1}^{-2}+{{\mathrm{t}}_2}^{-2}$

3. ${{\mathrm{t}}_0}^{-1}={{\mathrm{t}}_1}^{-1}+{{\mathrm{t}}_2}^{-1}$

4. ${\mathrm{t}}_0={\mathrm{t}}_1+{\mathrm{t}}_2$